Tính chất đường trung trực của một đoạn thẳng. Tính chất ba đường trung trực của một tam giác-level2

Home Lớp 7 Toán lớp 7 - Sách kết nối tri thức Tính chất đường trung trực của một đoạn thẳng. Tính chất ba đường trung trực của một tam giácBài tập nâng cao

{"common":{"save":0,"post_id":"1515","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"1971","post_id":"1515","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["160"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho $\\widehat{xOy} = 80^{o}$, \u0111i\u1ec3m $A$ n\u1eb1m trong $\\widehat{xOy}$. V\u1ebd \u0111i\u1ec3m $B$ sao cho $Ox$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$. V\u1ebd \u0111i\u1ec3m $C$ sao cho $Oy$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AC$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $BOC$. <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{BOC} = $ _input_ $^o$ ","hint":"Ch\u1ee9ng minh $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ r\u1ed3i t\u00ednh $\\widehat{BOC}$","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ <br\/> B\u01b0\u1edbc 2: <\/b> T\u00ednh $\\widehat{BOC}$ d\u1ef1a v\u00e0o g\u00f3c $xOy$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B22_K05.png' \/><\/center> <br\/> G\u1ecdi $I; K$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AB$ v\u1edbi $Ox$ v\u00e0 $AC$ v\u1edbi $Oy$ <br\/> V\u00ec $Ox$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ (gt) <br\/> $\\Rightarrow$ $OA = OB$ (\u0111\u1ecbnh l\u00fd thu\u1eadn - t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c) (1) <br\/> $Oy$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AC$ (gt) <br\/> $\\Rightarrow$ $OA = OC$ (\u0111\u1ecbnh l\u00fd thu\u1eadn - t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c) (2) <br\/> T\u1eeb (1) v\u00e0 (2) $\\Rightarrow$ $OB = OC$ $\\Rightarrow$ $O$ thu\u1ed9c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $BC$ (\u0111\u1ecbnh l\u00fd \u0111\u1ea3o - t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng trung tr\u1ef1c) <br\/> $\\triangle{OAB}$ c\u00f3 $OA = OB$ n\u00ean $\\triangle{OAB}$ c\u00e2n t\u1ea1i $O$ <br\/> $\\Rightarrow$ $Ox$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{AOB}$ (T\u00ednh ch\u1ea5t trung tr\u1ef1c trong tam gi\u00e1c c\u00e2n) <br\/> $\\Rightarrow$ $ \\widehat{O_{1}} = \\widehat{O_{2}}$ <br\/> T\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 $\\triangle{OAC}$ c\u00e2n t\u1ea1i $C$ $\\Rightarrow$ $\\widehat{O_{3}} = \\widehat{O_{4}}$ <br\/> $\\begin{align} \\widehat{BOC} &= \\widehat{O_{1}} + \\widehat{O_{2}} + \\widehat{O_{3}} + \\widehat{O_{4}} \\\\ &= \\widehat{O_{2}} + \\widehat{O_{2}} + \\widehat{O_{3}} + \\widehat{O_{3}} \\\\ &= 2\\widehat{O_{2}} + 2\\widehat{O_{3}} \\\\ &= 2(\\widehat{O_{2}} + \\widehat{O_{3}}) \\\\ &= 2 \\widehat{xOy} \\\\ &= 2 . 80^{o} \\\\ &= 160^{o} \\end{align}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $160$ <\/span> <br\/> Nh\u1eadn x\u00e9t: +) Trong m\u1ed9t tam gi\u00e1c c\u00e2n, \u0111\u01b0\u1eddng trung tr\u1ef1c \u1ee9ng v\u1edbi c\u1ea1nh \u0111\u00e1y \u0111\u1ed3ng th\u1eddi l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c, \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u00e0 \u0111\u01b0\u1eddng cao c\u00f9ng xu\u1ea5t ph\u00e1t t\u1eeb \u0111\u1ec9nh \u0111\u1ed1i di\u1ec7n v\u1edbi c\u1ea1nh \u0111\u00f3 <br\/> +) Trong tam gi\u00e1c \u0111\u1ec1u \u0111\u01b0\u1eddng cao c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn, \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c, \u0111\u01b0\u1eddng trung tr\u1ef1c <\/i> <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:06"},{"id":"1972","post_id":"1515","mon_id":"0","chapter_id":"0","question":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","options":{"time":24,"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n \u0111\u00fang v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["50"]]],"list":[{"point":10,"width":50,"type_input":"","ques":" Cho tam gi\u00e1c $ABC$ c\u00f3 $\\widehat{BAC} = 115^{o}$. C\u00e1c \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a $AB$ v\u00e0 $AC$ l\u1ea7n l\u01b0\u1ee3t c\u1eaft c\u1ea1nh $BC$ \u1edf $Q$ v\u00e0 $N$. T\u00ednh s\u1ed1 \u0111o g\u00f3c $QAN$ <br\/> \u0110\u00e1p \u00e1n l\u00e0: <\/b> $\\widehat{QAN} = $ _input_ $^{o}$ ","hint":"","explain":"H\u01b0\u1edbng d\u1eabn:<\/span><br\/> B\u01b0\u1edbc 1: <\/b> Ch\u1ee9ng minh $\\widehat{B} = \\widehat{BAQ}$ <br\/> B\u01b0\u1edbc 2: <\/b>Ch\u1ee9ng minh $\\widehat{C} = \\widehat{CAN}$ <br\/> B\u01b0\u1edbc 3: <\/b> T\u00ednh $\\widehat{BAQ} + \\widehat{CAN}$ <br\/> B\u01b0\u1edbc 4: <\/b> T\u00ednh $\\widehat{QAN}$ <br\/><br\/> B\u00e0i gi\u1ea3i:<\/span><br\/> <center><img src='img\/H7C3B22_K14.png' \/><\/center> <br\/> G\u1ecdi $P$, $M$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB; AC$ <br\/> Khi \u0111\u00f3, $PQ$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AB$ $\\Rightarrow$ $QA = QB$ (t\u00ednh ch\u1ea5t) <br\/> $MN$ l\u00e0 trung tr\u1ef1c c\u1ee7a $AC$ $\\Rightarrow$ $NA = NC$ (t\u00ednh ch\u1ea5t) <br\/> $\\blacktriangleright$ X\u00e9t hai tam gi\u00e1c vu\u00f4ng $QAP$ v\u00e0 $QBP$ c\u00f3: <br\/> $\\begin{cases} PQ \\hspace{0,2cm} \\text{chung} \\\\ QA = QB (cmt) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{QAP} = \\triangle{QBP}$ (c\u1ea1nh huy\u1ec1n, c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{B} = \\widehat{A_{1}}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (1) <br\/> Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $\\triangle{NAM} = \\triangle{NCM}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{A_{3}} = \\widehat{C}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) (2) <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3: $\\widehat{BAC} + \\widehat{B} + \\widehat{C} = 180^{o}$ (t\u1ed5ng ba g\u00f3c trong $1$ tam gi\u00e1c) <br\/> $ \\begin{align} \\Rightarrow \\widehat{B} + \\widehat{C} &= 180^{o} - \\widehat{BAC} \\\\ &= 180^{o} - 115^{o} \\\\ &= 65^{o} (3) \\end{align}$ <br\/> T\u1eeb (1), (2), (3) $\\Rightarrow$ $\\widehat{A_{1}} + \\widehat{A_{3}} = 65^{o}$ <br\/> $\\blacktriangleright$ M\u1eb7t kh\u00e1c ta c\u00f3: $\\widehat{A_{1}} + \\widehat{A_{2}} + \\widehat{A_{3}} = \\widehat{BAC}$ <br\/> $\\begin{align} \\Rightarrow \\widehat{A_{2}} &= \\widehat{BAC} - (\\widehat{A_{1}} + \\widehat{A_{3}}) \\\\ &= 115^{o} - 65^{o} \\\\ &= 50^{o} \\end{align}$ <br\/> Hay $\\widehat{QAN} = 50^{o}$ <br\/> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0: $50$ <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:06"},{"id":"1973","post_id":"1515","mon_id":"0","chapter_id":"0","question":"","options":{"time":24,"part":[{"title":"","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","4"]],"list":[{"point":10,"img":"","ques":"H\u00e3y ch\u1ecdn nh\u1eefng \u0111\u00e1p \u00e1n \u0111\u00fang.","hint":"","column":1,"number_true":2,"select":["A. \u0110\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a m\u1ed9t \u0111o\u1ea1n th\u1eb3ng l\u00e0 t\u1eadp h\u1ee3p t\u1ea5t c\u1ea3 c\u00e1c \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u hai \u0111\u1ea7u m\u00fat c\u1ee7a \u0111o\u1ea1n th\u1eb3ng \u0111\u00f3. ","B. Trong tam gi\u00e1c c\u00e2n, tr\u1ecdng t\u00e2m v\u00e0 \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u ba \u0111\u1ec9nh tr\u00f9ng nhau","C. \u0110\u01b0\u1eddng tr\u00f2n ngo\u1ea1i ti\u1ebfp tam gi\u00e1c c\u00f3 t\u00e2m l\u00e0 giao \u0111i\u1ec3m ba ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c \u0111\u00f3.","D. Trong tam gi\u00e1c \u0111\u1ec1u, ba \u0111\u01b0\u1eddng trung tr\u1ef1c c\u0169ng \u0111\u1ed3ng th\u1eddi l\u00e0 ba \u0111\u01b0\u1eddng trung tuy\u1ebfn, ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c."],"explain":" A - \u0110\u00daNG <\/b> theo Nh\u1eadn x\u00e9t <\/b> s\u00e1ch gi\u00e1o khoa To\u00e1n 7 - trang 75 <br\/> B - SAI v\u00ec: <\/b> <br\/> V\u00ed d\u1ee5 nh\u01b0 h\u00ecnh v\u1ebd d\u01b0\u1edbi \u0111\u00e2y, $\\triangle{ABC}$ c\u00e2n t\u1ea1i $A$ nh\u01b0ng tr\u1ecdng t\u00e2m $G$ kh\u00e1c \u0111i\u1ec3m $I$ l\u00e0 \u0111i\u1ec3m c\u00e1ch \u0111\u1ec1u ba \u0111\u1ec9nh c\u1ee7a tam gi\u00e1c <br\/> <center><img src='img\/H7C3B22_K01.png' \/><\/center> <br\/> C - Sai v\u00ec: <\/b> <br\/> <center><img src='img\/H7C3B22_K01C.png' \/><\/center> <br\/> $\\triangle{ABC}$ c\u00f3 $O$ l\u00e0 giao \u0111i\u1ec3m ba \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c n\u00ean ta c\u00f3: <br\/> $OH = OK = OM$ (t\u00ednh ch\u1ea5t) <br\/> $\\Rightarrow$ $O$ c\u00e1ch \u0111\u1ec1u ba c\u1ea1nh c\u1ee7a tam gi\u00e1c hay $O$ l\u00e0 t\u00e2m \u0111\u01b0\u1eddng tr\u00f2n n\u1ed9i ti\u1ebfp tam gi\u00e1c <br\/> D - \u0110\u00daNG v\u00ec: <\/b> <br\/> <center><img src='img\/H7C3B22_K01D.png' \/><\/center> <br\/> Gi\u1ea3 s\u1eed $\\triangle{ABC}$ \u0111\u1ec1u c\u00f3 $AM; CN; BP$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $BC; AB$ v\u00e0 $AC$. Ta c\u1ea7n ch\u1ee9ng minh $AM, CN, BP$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c <br\/> $\\blacktriangleright$ $\\triangle{ABC}$ c\u00f3: $AM$ l\u00e0 \u0111\u01b0\u1eddng trung tr\u1ef1c c\u1ee7a c\u1ea1nh $BC$ <br\/> N\u00ean $BM = MC; AB = AC$ (t\u00ednh ch\u1ea5t) <br\/> V\u00ec $BM = MC$ n\u00ean $AM$ c\u0169ng l\u00e0 \u0111\u01b0\u1eddng trung tuy\u1ebfn \u1ee9ng v\u1edbi c\u1ea1nh $BC$ <br\/> X\u00e9t hai tam gi\u00e1c vu\u00f4ng $AMB$ v\u00e0 $AMC$ c\u00f3: <br\/> $\\begin{cases} AM \\hspace{0,2cm} \\text{chung} \\\\ AB = AC (\\triangle{ABC} \\hspace{0,2cm} \\text{\u0111\u1ec1u}) \\end{cases}$ <br\/> $\\Rightarrow$ $\\triangle{AMB} = \\triangle{AMC}$ (c\u1ea1nh huy\u1ec1n - c\u1ea1nh g\u00f3c vu\u00f4ng) <br\/> $\\Rightarrow$ $\\widehat{BAM} = \\widehat{CAM}$ (hai g\u00f3c t\u01b0\u01a1ng \u1ee9ng) <br\/> $\\Rightarrow$ $AM$ l\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a $\\widehat{BAC}$ <br\/> $\\blacktriangleright$ Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1 ta c\u0169ng c\u00f3 $BN; CP$ c\u0169ng l\u00e0 \u0111\u1ed3ng th\u1eddi \u0111\u01b0\u1eddng trung tuy\u1ebfn v\u00e0 \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c <br\/> V\u1eady c\u00e1c \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0: A; D <\/span> "}]}]},"correct":"","level":"3","hint":"","answer":"","type":"json","extra_type":"","time":"0","user_id":"0","test":"0","date":"2019-09-30 09:23:06"}]}